Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s)
a. 1 mol Al, 1 mol O₂
b. 4 mol Al, 2.6 mol O₂
c. 16 mol Al, 13 mol O₂
d. 7.4 mol Al, 6.5 mol O₂

Question

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s)

a. 1 mol Al, 1 mol O₂

b. 4 mol Al, 2.6 mol O₂

c. 16 mol Al, 13 mol O₂

d. 7.4 mol Al, 6.5 mol O₂

Accepted Answer

Hello, everyone. Today we have the following problem. Write the balanced chemical equation for each reaction. So we have B Cobalt three oxide reacts with hydrogen gas to form solid Cobalt and liquid water. So what does that look like? So we would have our Cobalt three oxide which is a solid reacting with hydrogen gas, which is of course, a gas and this will react to form solid Cobalt and liquid water. So we look at this equation and we see that it is not quite balanced. Of course, if we compare Adams on both sides, we can see that on the left, we have two Cobalts, but on the right, we have one on the left, we have three oxygens, but on the right, we have two, we have one and then hydrogens, we have two on the left and two on the right. So we need to balance the Cobalts to do that. We need to add a two to the coefficient of cobalt. And our equation, we multiply that one by the two to get two. And the next we balance our oxygens. We see that we have three on the left and one on the right. So we need to add a coefficient of three to our water. Such that when we multiply that oxygen by three, we have three, but we also change the number of hydrogens. So instead of two, we now have three times two or six. So what number can we multiply our hydrogen gas on the left to get to six, three. So if you multiply that by three, we get six and now it's balanced. And if we look at our anti choices, we see that choice c best reflects this and with that, we have solved the problem overall, I hope this helped. And until next time.

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s)
a. 1 mol Al, 1 mol O₂
b. 4 mol Al, 2.6 mol O₂
c. 16 mol Al, 13 mol O₂
d. 7.4 mol Al, 6.5 mol O₂

Verified Solution

Key Concepts

Limiting Reactant

Stoichiometry

Mole Ratio

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O2( g) → 2 Al2O3(s)a. 1 mol Al, 1 mol O2b. 4 mol Al, 2.6 mol O2c. 16 mol Al, 13 mol O2d. 7.4 mol Al, 6.5 mol O2

Verified Solution

Key Concepts

Limiting Reactant

Stoichiometry

Mole Ratio

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s)
a. 1 mol Al, 1 mol O₂
b. 4 mol Al, 2.6 mol O₂
c. 16 mol Al, 13 mol O₂
d. 7.4 mol Al, 6.5 mol O₂